Question: Find $\dfrac{d}{dx}\left(e^x\sin(x)\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $e^x(\sin(x)+\cos(x))$ (Choice B) B $e^x(\sin(x)-\cos(x))$ (Choice C) C $e^x\cos(x)$ (Choice D) D $-e^x\cos(x)$
Solution: $e^x\sin(x)$ is the product of two, more basic, expressions: $e^x$ and $\sin(x)$. Therefore, the derivative of the expression can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(e^x\sin(x)\right) \\\\ &=\dfrac{d}{dx}\left(e^x\right)\sin(x)+e^x\dfrac{d}{dx}(\sin(x))&&\gray{\text{The product rule}} \\\\ &=e^x\cdot \sin(x)+e^x\cdot \cos(x)&&\gray{\text{Differentiate }e^x\text{ and }\sin(x)} \\\\ &=e^x(\sin(x)+\cos(x))&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(e^x\sin(x)\right)=e^x(\sin(x)+\cos(x))$